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The triangles have sides $3,3,1$, or $3,2,2$. You can compute the area of each using Heron's Formula.

Or else note these are both isosceles. The height of the $3,3,1$, with respect to base $1$, is $\sqrt{3^2-\frac{1}{4}}$ (Pythagorean Theorem). You can now find its area. Do a similar calculation for the other triangle.

Note: The area of an isosceles triangle whose sides are known is easily obtained using Pythagoras (cut the triangle along the axis of symmetry and compute the height directly) if Heron's formula is not to hand.
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Mark BennetAug 31 '13 at 7:15

But, is there a general method for doing these types of problems.What it the sum of sides is 75? In such cases writing all the possibilities will not work were time matters.
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Rajath Krishna RAug 31 '13 at 7:17

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Then it will be the equilateral triangle. In general the most efficient way to enclose area is to have the sides as near each other as possible.
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André NicolasAug 31 '13 at 7:19

To see that the area is maximum when sides are close fix the sum of the sides $2s=a+b+c$ and note that the product $s(s-a)(s-b)(s-c)$ which appears in Heron's formula is greatest when $(s-a)(s-b)(s-c)$ is greatest. The sum of these three terms is $s$. The product of terms having a fixed sum is greatest when the terms are equal -$AM\ge GM$
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Mark BennetAug 31 '13 at 7:45